从for循环Python打印数字列表作为数组

Posted 2021-03-28

使用下面的代码，它会逐个打印值“phase”。我试图将这些值打印为for循环外的数组。

import math

Period = 6.2

time1 = datafile1[:,0]
magnitude1 = datafile1[:,1]
for i in range(len(time1)):
   print(i,time1[i])
   floor = math.floor((time1[i]-time1[0])/Period)
   phase = ((time1[i]-time1[0])/Period)-floor 
   print (phase)

像这样打印：

0.002
0.003
0.004
0.005

我希望它像这样打印：

[0.002, 0.003, 0.004, 0.005]

答案

这将是该结果的最小修改要求路径

result = []

time1 = datafile1[:,0]
magnitude1 = datafile1[:,1]
for i in range(len(time1)):
   result.append(i,time1[i])
   floor = math.floor((time1[i]-time1[0])/Period)
   phase = ((time1[i]-time1[0])/Period)-floor 
   result.append(phase)

print(result)

另一答案

在这里，我已经做到了，而不是printing您的结果，您将它们附加到列表，然后打印出完整列表。

import math

Period = 6.2

time1 = datafile1[:,0]
magnitude1 = datafile1[:,1]

my_list = []
for i in range(len(time1)):
   my_list.append(i,time1[i])
   floor = math.floor((time1[i]-time1[0])/Period)
   phase = ((time1[i]-time1[0])/Period)-floor 
   my_list.append(phase)

print(my_list)

另一答案

你可以做

import math

Period = 6.2

time1 = datafile1[:,0]
magnitude1 = datafile1[:,1]
list_to_print = []
for i in range(len(time1)):
   print(i,time1[i])
   floor = math.floor((time1[i]-time1[0])/Period)
   phase = ((time1[i]-time1[0])/Period)-floor
   list_to_print.append(phase)
print (list_to_print)